1. The explanatory variable is the number of cans of beer. The response variable is the blood alcohol level.
2. There are two explanatory variables: amount of debt and income. The response variable is stress caused by college debt.
1. The relationship is positive. The longer the duration of the eruption, the longer the wait between eruptions is. One reason for this may be that if the geyser erupted for longer, it expended more energy and it will take longer to build up the energy needed to erupt again.
2. The form is roughly linear with two clusters. The clusters indicate that in general there are two types of eruptions, one shorter, the other somewhat longer.
3. The relationship is fairly strong. Two points define a line and in this case we could think of each cluster as a point, so the two clusters seem to define a line.
4. There are a few outliers around the clusters, but not many and not very distant from the main grouping of points.
5. The Starnes family needs to know how long the last eruption was in order to predict how long it will be until the next one.
1. (a) The correlation is about 0.9. This indicates that there is a strong, positive linear relationship between the number of boats registered in Florida and the number of manatees killed. (b) The correlation is about 0.5. This indicates that there is a moderate, positive linear relationship between the number of named storms predicted and the actual number of named storms. (c) The correlation is about 0.3. This indicates that there is a weak, positive linear relationship between the healing rate of the two front limbs of the newts. (d) The correlation is about -0.1. This indicates that there is a weak, negative linear relationship between last year’s percent return and this year’s percent return in the stock market.
2. If we remove the outlier in this scatterplot, the correlation would decrease. This point has the effect of strengthening the observed linear relationship we see.
3.1 Water temperature is the explanatory variable, and weight change (growth) is the response variable.
Both are quantitative.
3.2 The explanatory variable is the type of treatment—removal of the breast or removal of only the tumor and nearby lymph nodes, followed by radiation, and survival time is the response variable. Type of treatment is a categorical variable, and survival time is a quantitative variable.
3.3 (a) A positive association between IQ and GPA means that students with higher IQs tend to have higher GPAs, and those with lower IQs generally have lower GPAs. The plot does show a positive association. (b) The form of the relationship is roughly linear, because a line through the scatterplot of points would provide a good summary. The positive association is moderately strong (with a few exceptions) because most of the points would be close to the line. (c) The lowest point on the plot is for a student with an IQ of about 103 and a GPA of about 0.4.
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3.4 (a) The plot shows a negative association. This is what we would expect because as the temperature gets warmer, she would not need to heat her house as much, so her use of gas would decline. (b) The scatterplot shows a strong linear relationship. It is linear because a line through the scatterplot of points would provide a good summary and it is strong because the points would all be close to the line. (c) The point in the bottom right of the plot represents a month when the average temperature was about 58 degrees and the gas usage was about 260 cubic feet.
3.5 A scatterplot is shown below.
3.6 A scatterplot is shown below.
3.7 (a) The scatterplot shows a positive, somewhat curved, moderately weak association. (b) The outlier is Hawaiian Airlines. Without this outlier, the relationship is more linear but still not very strong.
3.8 (a) The scatterplot shows a negative, linear, fairly strong relationship. (b) Because this association is negative, we conclude that the sparrowhawk is a long-lived territorial species.
Chapter 3: Describing Relationships 63

3.9 (a) A scatterplot with speed as the explanatory variable is shown below.
(b) The relationship is curved or quadratic. High amounts of fuel were used for low and high values of speed and low amounts of fuel were used for moderate speeds. This makes sense because the best fuel efficiency is obtained by driving at moderate speeds. (Note: 60 km/hr is about 37 mph) (c) Both are present. The first part of the graph (low speeds) would be described as negative and the second part
(higher speeds) would be positive. (d) The relationship is very strong, with little deviation for a curve that can be drawn through the points.
3.10 (a) A scatterplot with mass as the explanatory variable is shown below.
(b) The association is positive, and the relationship is linear and moderately strong.
3.11 (a) Most of the southern states blend in with the rest of the country. Several southern states do lie at the lower edges of their clusters. This means that, in general, the students in the southern states do not do as well as their counterparts in other portions of the country. (b) West Virginia is an outlier because it has a much lower mean SAT Math score than the other states which have a similar percent of students taking the exam.
3.12 The scatterplot shows that the pattern of the relationship does hold both for men and women.
However, the relationship between mass and rate is not as strong for men as it is for women. The group of men has higher lean body masses and metabolic rates than the group of women.
3.13 State: Is the relationship between the number of breeding pairs of merlins and the percent of males who return the next season negative? Plan: We will begin with a scatterplot, and compute the correlation if appropriate. Do: A scatterplot of the percent returning against the number of breeding pairs (shown below) shows the expected negative association. Though slightly curved, it is reasonable to compute
64 The Practice of Statistics for AP* , 4/e

r = − 0.7943
as a measure of the strength of the linear association. Conclude: This supports the theory: a smaller percent of birds survive following a successful breeding season.
3.14 State: Does social rejection cause activity in areas of the brain that are known to be activated by physical pain? Plan: We will begin with a scatterplot, and compute the correlation if appropriate. Do:
A scatterplot (shown below) shows a fairly strong positive linear association. There are no particular outliers; each variable has low and high values, but those points do not deviate from the pattern of the rest. The relationship seems to be reasonably linear, so we compute increased activity in the pain-sensing area of the brain. r = 0.8782
. Conclude: Social exclusion does appear to trigger a pain response: higher social distress measurements are associated with
3.15 (a) r = 0.9 (b) r = 0 (c) r = 0.7 (d) r = − 0.3
(e) r = − 0.9
3.16 Answers may vary. We would expect the height of women at age 4 and their height as women at age 18 to be the highest correlation since it is reasonable to expect taller children to become taller adults and shorter children to become shorter adults. The next highest would be the correlation between the heights of male parents and their adult children. Tall fathers tend to have tall sons, but typically not as tall, and likewise for shorter fathers. The lowest correlation would be between husbands and their wives.
Husbands may be taller than their wives in general, but there is no reason to expect anything more than a weak positive correlation.
3.17 (a) Gender is a categorical variable and the correlation coefficient r measures the strength of linear association for two quantitative variables. (b) The largest possible value of the correlation coefficient r is
1. (c) The correlation coefficient r has no units.
3.18 The paper’s report is wrong because the correlation ( r = 0.0) is interpreted incorrectly. The author incorrectly suggests that a correlation of zero indicates a negative association between research productivity and teaching rating. The psychologist meant that there is no linear association between
Chapter 3: Describing Relationships 65

research productivity and teaching rating. In other words, knowledge of a professor’s research productivity will not help you predict her teaching rating.
3.19 (a) The scatterplot below shows a strong, positive, linear relationship between the two measurements. Thus, all five specimens appear to be from the same species. femur
38
56
59
64
74
Humerus
41
63
70
72
84 zfemur
-1.53048
-0.16669
0.06061
0.43944
1.19711 zhumerus
-1.57329
-0.18880
0.25173
0.37759
1.13277 product
2.40789
0.03147
0.01526
0.16593
1.35605
(b) The femur measurements have mean of 58.2 and a standard deviation of 13.2. The humerus measurements have a mean of 66 and a standard deviation of 15.89. The table below shows the standardized measurements (labeled zfemur and zhumerus) obtained by subtracting the mean and dividing by the standard deviation. The column labeled “product” contains the product (zfemur×zhumerus) of the standardized measurements. The sum of the products is 3.97659, so the correlation coefficient is r =
4
( ) = .
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3.20 (a) The scatterplot shows a moderate positive association, so r should be positive, but not close to 1.
(b) For the women, the mean is 66 and the standard deviation is 2.098. For the men, the mean is 69 and the standard deviation is 2.53. The table below shows the standardized measurements (labeled zfemale and zmale) obtained by subtracting the mean and dividing by the standard deviation. The column labeled
“product” contains the product (zfemale×zmale) of the standardized measurements. The sum of the products is 2.82667, so the correlation coefficient is r =
( )
. female
66
64
66
65
70
65 male
72
68
70
68
71
65 zfemale
0 zmale
1.18585 product
0
-0.95346 -0.39528 0.37689
0 0.39528 0
-0.47673 -0.39528 0.18844
1.90693 0.79057 1.50756
-0.47673 -1.58114 0.75378
There is some evidence that taller women tend to date taller men (and shorter women date shorter men), but it is hardly overwhelming---and the small sample size makes any conclusion suspect.
3.21 (a) There is a strong positive linear association. High-calorie hot dogs tend to be high in salt, and low-calorie hot dogs tend to have lower sodium. (b) It would tend to decrease the correlation. An outlier generally in line with the bulk of the data will tend to increase the correlation.
3.22 (a) There is a strong positive linear association between body weight and brain weight of mammals.
(b) It would tend to decrease the correlation. An outlier generally in line with the bulk of the data will tend to increase the correlation.
3.23 (a) The correlation would not change. It does not have units associated with it, so a change in units for either variable (or both) will not change the correlation. Multiplying both the x and y values by 10 will also multiply their standard deviations by 10, so the z -scores will not change. (b) The correlation would not change. The correlation measures the strength of the linear relationship between two quantitative variables. It does not distinguish between the explanatory and response variables.
3.24 (a) If all the men were 6 inches shorter, the correlation would not change. The correlation tells us that there is a weak to moderate association between women’s heights and men’s heights (that is, that taller women tend to date taller men), but it does not tell us whether or not they tend to date men taller than themselves. Subtracting 6 from each y would also subtract 6 from the mean so the z -scores would
Chapter 3: Describing Relationships 67

remain unchanged. (b) The correlation would not change because correlation does not have units associated with it.
3.25 The scatterplot is shown below. The one unusual point (10, 1) is responsible for reducing the correlation. Outliers tend to have fairly strong effects on correlation; the effect is very strong here because there are only six observations.
3.26 A scatterplot of mileage versus speed is shown below.
The correlation coefficient r measures the strength of linear association between two quantitative variables; this plot shows a nonlinear relationship between speed and mileage.
3.27 a
3.28 e
3.29 d
3.30 b
3.31 c
3.32 d
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3.33 Use either a histogram or a stemplot. The histogram is shown below. The distribution is sharply right-skewed, with several possible high outliers. The five number summary is 0.1, 3.5, 5.4, 9, 33.8.
3.34 (a) The mean
M Q
1
= $5,632 x and
=
Q
$17,776
3
− = so
3
Meanwhile,
Q is further from the median than
1 comparisons result in what we would expect for right-skewed distributions. (b) From Table A, we estimate that the third quartile of a Normal distribution would be 0.675 standard deviations above the mean, which would be ( ) = $25,899.
(Software gives 0.6745, which yields
$25,893.) As the exercise suggests, this quartile is larger than the actual value of
1. The slope is 40. We predict that a rat will gain 40 grams of weight per week.
2. The y -intercept is 100. This suggests that we expect a rat at birth to be 100 grams.
3. After 16 weeks, we predict the rat’s weight to be y ( ) grams.
4. The time is measured in weeks for this equation, so 2 years becomes 104 weeks. We then predict the rat’s weight to be y ( ) grams which is equivalent to 9.4 pounds (about the weight of a large newborn human). This is unreasonable and is the result of extrapolation.
1. The answer is given in the text.
1. We predict the fat gain for this person to be ˆ 3.505 0.00344(620) 1.3722
kg. So their residual is
2. The residual says that this person gained 0.9278 kg more than we would have predicted using the least-squares line as a model.
3. The line over predicted the fat gain the most for the person who had an NEA change of 580 and a fat gain of 0.4. Their predicted fat gain would be 1.51 so their residual is -1.11. Based on the list of residuals given in the previous example, this is the largest negative residual, and therefore the point for which the fat gain was most over predicted.
Chapter 3: Describing Relationships 69

1. There is a moderate, positive linear relationship with one outlier in the bottom right corner of the plot.
2. The average error (residual) in predicting the backpack weight is 2.27 using the least-squares regression line.
1. c.
2. d.
3.35 The equation is ˆ 80 6 x where ˆ y = the estimated weight of the soap and x = the number of days since the bar was new.
3.36 The equation is ˆ 50 + x where ˆ y = the predicted reading test score and x = the number of points above 100 for a child’s IQ (this would be negative for a child whose IQ is less than 100).
3.37 (a) The slope is 1.109. We predict highway mileage will increase by 1.109 mpg for each 1 mpg increase in city mileage. (b) The intercept is 4.62 mpg. This is not statistically meaningful because this would represent the highway mileage for a car that gets 0 mpg in the city. (c) With city mpg of 16, the predicted highway mpg is 4.62 1.109(16) 22.36
+ = mpg.
mpg. With city mpg of 28, the predicted highway
3.38 (a) The slope is 0.882; this means that we predict reading scores will increase by 0.882 for each onepoint increase in IQ. (b) The y-intercept is -33.4. This would only be statistically meaningful if a child could have an IQ score of 0. (c) The predicted scores for 90
− 33.4 0.882(90) 45.98
x = and x = 130 are
3.39 (a) The slope is −0.0053; this means that on the average for each additional week in the study the pH decreased by 0.0053 units. Thus, the acidity of the precipitation increased over time. (b) The y intercept is 5.43 and it provides an estimate for the pH level at the beginning of the study. (c) At the end of the study pH is predicted to be 5.43 0.0053(150) 4.635.
3.40 (a) The slope is -19.87. We predict the amount of gas consumed in Joan’s home to decrease by
19.87 cubic feet for every degree the average monthly temperature increases. (b) The y -intercept is 1425.
When the average monthly temperature is 0 ° F, the predicted gas consumption for Joan’s home is 1425 cubic feet. This is an extrapolation since the data only included points for months with an average temperature of more than 20 ° F. (c) cubic feet. We predict that the amount of natural gas Joan will use in a month with an average temperature of 30 ° F is 828.9 cubic feet.
3.41 No. The data was collected weekly for 150 weeks. 1000 months corresponds to roughly 4000 weeks which is well outside the observed time period. We do not know that the linear relationship continues after 150 weeks. This constitutes extrapolation.
3.42 No. The average temperatures for the months where data was collected were between about 27 ° F and 57 ° F. 65 ° F is outside of this range so using the line to make a prediction here would be considered extrapolation. We do not know that the linear relationship continues after 57 ° F.
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3.43 The dotted (red) line is the line ˆ 1 x y = − x The dotted line comes closer to all of the data points. Thus, the line ˆ 1 x fits the data best.
3.44
This line minimizes the square of the vertical distance between the points and the line.
So the residual is
5.08 5.185
large.
= − 0.085.
This means that the line predicted a pH value for that week that was 0.085 too
So the residual is
490 503.32
= − 13.032.
This means that the line predicted that Joan would use 13.032 cubic feet of gas per day more than she actually did.
3.47 (a) The slope is equation for predicting b y
= 0.5

 2.7


2.5


= 0.54.
The y
= husband’s height from
intercept is x a = − y = +
= x
So the
(b) The
inches. 67 inches is one standard deviation above the mean for women. So the predicted value for husband’s height would be y rs y
=
3.48 (a) The slope is b y
=
=
0.596



+
15.35
5.36


 x
= 1.707.
The y intercept is a =
(b) The predicted change is
− = So the
Chapter 3: Describing Relationships 71

y + = We could have given the answer without doing calculations because the regression line must pass through ( ) (
3.49 (a) r 2 = ( ) 2
= 0.25.
Thus, the straight-line relationship explains 25% of the variation in husbands heights. (b) The average error (residual) when using the line for prediction is 1.2 inches.
3.50 (a) r 2 =
( 0.596
) 2
= 0.3552.
Thus, the straight-line relationship explains 35.52% of the variation in yearly changes. (b) The average error (residual) when using the line for prediction is 8.3%.
3.51 (a) The least-squares line for predicting and intercept a = − = − y = GPA from
3.5519.
x = IQ has slope b = 0.6337



2.1
13.17



=
Thus, the regression line is ö = − + x
0.101
(b) r 2 =
(
0.6337
) 2
= 0.4016.
Thus, 40.16% of the variation in GPA is accounted for by the linear relationship with IQ. (c) The predicted GPA for this student is ˆ y = − + = and the residual is
0.53 6.8511
= − 6.3211.
This means that the student had a GPA that was 6.3211 points worse than expected for someone with an IQ of 103.
3.52 Since the least-squares regression line must pass through the point of averages, we know that y = Octavio’s predicted final exam score is y = + ( x + 10 ) ( + x ) + 0.41(10) y 4.1.
Thus, we predict that he will score 4.1 points above the mean on the final exam.
3.53 (a) The scatterplot is y = − x scatterplot above as well.
Minitab output
The regression equation is newadults = 31.9 - 0.304 %returning
Predictor Coef SE Coef T P
Constant 31.934 4.838 6.60 0.000
%returning -0.30402 0.08122 -3.74 0.003
S = 3.66689 R-Sq = 56.0% R-Sq(adj) = 52.0%
Minitab output is shown below. See the
72 The Practice of Statistics for AP* , 4/e

(c) The slope tells us that as the percent of returning birds increases by one, we predict the number of new birds will decrease by −0.304. The y intercept provides a prediction that we will see 31.9 new adults in a new colony when the percent of returning birds is zero. This value is clearly outside the range of values studied for the 13 colonies of sparrowhawks and has no practical meaning in this situation. (d)
The predicted value for the number of new adults is 31.9 0.304(60) 13.66
3.54 (a) The scatterplot is shown below.
or about 14. y = + x Minitab output is shown below. See the scatterplot above as well.
Minitab output
The regression equation is
Rate = 201 + 24.0 Mass
Predictor Coef SE Coef T P
Constant 201.2 181.7 1.11 0.294
Mass 24.026 4.174 5.76 0.000
S = 95.0808 R-Sq = 76.8% R-Sq(adj) = 74.5%
(c) The slope tells us that we would predict an increase in the metabolic rate of about 24 cal/day for each additional kilogram of body mass. (d) For x = 45 kg, the predicted metabolic rate is ˆ 1282.3 cal/day.
3.55 (a) The residual plot (shown below) suggests that the line is a decent fit. The points are all scattered around a residual value of 0. It is noteworthy that there are three larger negative residuals, but given the size of this data set, these are probably not too much of a concern.
(b) The point with the largest residual has a residual of about -6. This means that the line over-predicted the number of new adults by 6.
Chapter 3: Describing Relationships 73

3.56 (a) The residual plot (shown below) shows that the linear fit is good. There is one large, positive, outlier, but since it is near the mean of the mass values, it does not influence the line very much.
(b) The point that has the largest residual has a residual of about 200. This means that the line greatly under-predicted the metabolic rate for this particular person.
3.57 56% of the variation in the number of new adult birds is explained by the straight-line relationship.
The average error (residual) when using the line for prediction is 3.67%.
3.58 76.8% of the variation in the metabolic rate is explained by the straight-line relationship. The average error (residual) when using the line for prediction is 95.08 calories burned per 24 hours.
3.59 (a) There is a positive, linear association between the two variables. There is more variation in the field measurements for larger laboratory measurements. Also, the values are scattered above and below the line y = x for small and moderate depths, indicating strong agreement, but the field measurements tend to be smaller than the laboratory measurements for large depths. (b) The points for the larger depths fall systematically below the line = showing that the field measurements are too small compared to the laboratory measurements. (c) In order to minimize the sum of the squared distances from the points to the regression line, the top right part of the blue line in the scatterplot would need to be pulled down to go through the “middle” of the group of points that are currently below the blue line. Thus, the slope would decrease and the intercept would increase.
3.60 The residual plot clearly shows that the prediction errors increase for larger laboratory measurements. In other words, the variability in the field measurements increases as the laboratory measurements increase. The least squares line does not provide a great fit, especially for larger depths.
3.61 Clearly, this line does not fit the data very well; the data show a clearly curved pattern. The residual plot shows a clear curved pattern, with the first two and the last four residuals being negative and those between 3 and 8 months being positive.
3.62 We would certainly not use the regression line to predict fuel consumption. The scatterplot shows a nonlinear relationship.
Following a season with 30 breeding pairs, we find y − =
Minitab output as R-sq 63.1%.
returning males.
so we predict that about 68% of males will return. (b) This is given in the
The linear relationship explains 63.1% of the variation in the percent of
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(c) Knowing that r 2 = 0.631, we find s = r = − as the slope coefficient. (d) Since 9.46, of males is about 9.46%. r 2 = − 0.79;
3.64 (a) The regression equation is ˆ = − + x
the sign is negative because it has the same sign
the typical error when using the line to predict the return rate
For x = 2.0, this formula gives y ˆ = − + = − 0.0044.
The linear relationship explains 77.1% of the variation in brain activity. (c) Knowing that r 2 = 0.771, we find r = + s = r 2
0.0251,
= 0.88; the sign is positive because it has the same sign as the slope coefficient. (d) Since
the typical error when using the line to predict the activity in the brain is about 0.025.
3.65 (a) The scatterplot (with regression lines) is shown below.
(b) The correlation is is too small a change to consider the outlier influential for correlation. (c) With all points, y ˆ 4.73 0.3868
x r = 0.4765
with all points. It rises slightly to 0.4838 with the outlier removed; this
(the solid, blue, line), and the prediction for x = 76 is 34.13%. With Hawaiian
= + x (the dotted, black, line), and the prediction is 29.84%. This difference in prediction---and the visible difference in the two lines---indicates that the outlier is influential for regression.
3.66 (a) A scatterplot, with the two unusual observations marked and the three separate regression lines added, is shown below.
(b) The correlations are: r
1
= 0.4819
(all observations); r
2
= 0.5684
(without Subject 15); r
3
= 0.3837
(without Subject 18). Both outliers change the correlation. Removing subject 15 increases r , because its presence makes the scatterplot less linear, while removing Subject 18 decreases r , because its presence decreases the relative scatter about the linear pattern. (c) The three regression lines shown in the
Chapter 3: Describing Relationships 75

x x (without #15); y = + x (without #18). While the equation changes in response to removing either subject, one could argue that neither one is particularly influential, as the line moves very little over the range of x
(HbA) values. Subject #15 is an outlier in terms of its y value; such points are typically not influential.
Subject #18 is an outlier in terms of its x value, but is not particularly influential because it is consistent with the linear pattern suggested by the other points.
3.67 (a) A scatterplot with the two new points is shown below. Point A is a horizontal outlier; that is, it has a much smaller x -value than the others. Point B is a vertical outlier; it has a higher y value than the others. y = − (b) The three regression formulas are: ˆ 31.9 0.304
y = − x x (the original data, solid black line);
= − x (with Point B, dotted green line).
Adding Point B has little impact. Point A is influential; it pulls the line down, and changes how the line looks relative to the original 13 data points.
3.68 Answers may vary. For example: Weight, gender, other food eaten by the students, type of beer
(light, imported, …).
3.69 State: How accurate are Dr. Gray’s forecasts? Plan: We will construct a scatterplot with Dr. Gray's forecast as the explanatory variable, and if appropriate, find the regression equation. Then we should make a residual plot and calculate y = + r 2 and s . Do: The scatterplot shows a moderate positive association; x with r 2 = 0.30
and s = 4.0.
The relationship is strengthened by the large number of storms in the 2005 season, but it is weakened by the 2006 and 2007, when Gray's forecasts were the highest, but the actual numbers of storms were unremarkable. As an indication of the influence of the 2005 season, we might find the regression line without that point; it is y = + x with r 2 = 0.265
and s = 3.14.
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Finally, the residual plot does not indicate any problems with fitting the linear equation.
Conclude: If Dr. Gray forecasts x = 16 tropical storms, we expect 16.33 storms in that year. However, we do not have very much confidence in this estimate, because the regression line explains only 30% of the variation in tropical storms and the typical error we should expect when using this line for prediction is 4 storms. (If we exclude 2005, the prediction is 14.7 storms, but this estimate is less reliable than the first.)
3.70 State: Do more stumps result in more beetles? How accurate will our predictions be? Plan: We will construct a scatterplot with the number of stumps as the explanatory variable, and if appropriate, find the regression equation. Then we should make a residual plot and compute r 2 and s . Do: with the least-squares regression line, is shown below. The plot shows a strong, positive linear association between the number of beaver-caused stumps and the number of beetle larvae clusters.
A scatterplot,
Chapter 3: Describing Relationships 77

The least-squares regression line is ˆ = − + x model appears to provide a very good fit. and the residual plot is shown below. The linear
Finally, r 2 = 0.839
and s = 6.42.
In other words, about 84% of the variation in the number of beetle larvae clusters is accounted for by the linear relationship with the number of stumps and our average error in prediction is about 6.4 larvae. Conclude: More stumps (i.e. more beavers) does appear to lead to more beetle larvae. However we should be cautious because we have few observations with 4 or 5 stumps.
3.71 b
3.72 c
3.73 b
3.74 a
3.75 b
3.76 a
3.77 d
3.78 a
3.79 About 92.92%. For the N(18.7,4.3) distribution, x < 25 corresponds to z <
3.80 At least 24.2 mpg. Search Table A for the proportion closest to 0.90; this is z = 1.28, the 90th percentile for the N(0,1) distribution. The top 10% of all vehicles are those with gas mileage at least 1.28 standard deviations above the mean:
+ = mpg or more.
3.81 (a) There is evidence of an association between accident rate and marijuana use. Those people who use marijuana more are more likely to have caused accidents.
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(b) This was an observational study. All we can observe is that there is an association between these two variables. If we wanted to see whether using marijuana more caused more accidents, then we would have to set up an experiment where we randomly assigned people to use more or less marijuana. This is unethical.
(page 198)
R3.1 (a) The explanatory variable is weight of a person, and the response variable is mortality rate (that is, how likely a person is to die over a 10-year period). (b) We cannot conclude that increased weight causes a greater risk of dying because this was an observational study.
R3.2 (a) The direction of the scatterplot is positive, but appears curved, not linear. The strength of the association is moderate. (b) The hippopotamus is unusual because its lifespan is longer than would be expected given its gestation period, based on the association in the remaining data. The Asian elephant is unusual because it has the second to largest gestation time. The two animals with the largest gestation times do not follow the curvilinear patter seen in the shorter gestation times, rather, they tend to have much longer lifetimes as well. The giraffe’s observation tends to follow the curvilinear shape, with possibly a little shorter lifespan than is to be expected based on the pattern in the remaining data.
R3.3 (a) The slope is 0.0138 minutes per meter. We predict that if the depth of the dive is increased by one meter, it will add 0.0138 minutes (about 0.83 seconds) to the time spent underwater. (b) When Depth y = = minutes (5 minutes and 27 seconds). (c) The intercept suggests that a dive of no depth would last an average of 2.69 minutes; this obviously does not make any sense.
+ y where y represents the mileage of the cars and x represents the age. (b) For a 6-year old car, we predict a mileage of
The residual for this particular car is 65000 77072.14
= − 12072.14.
In other words, this teacher has driven 12,072.14 fewer miles than we would have predicted with the model. (c) The slope of the line is 11,630.6. We expect that cars will be driven 11,630.6 miles per year on average. (d) Since r 2 = 0.82
and the slope is positive, the correlation r = + = This shows that there is a strong linear relationship between the age of cars and their mileage. (e) The line fits reasonably well. The residual plot shows no large pattern, though there are three cars with large positive
Chapter 3: Describing Relationships 79

residuals (meaning they have been driven more than one would expect for their age). The average error
(residual) when using this line to predict mileage is 19,280. This suggests that, while the line fits reasonably well, it would not be very useful in practice since our predictions are going to be off by an average of approximately 20,000 miles.
R3.5 (a) The association is negative, the number of days in April until the first bloom decreases as the average March temperature increases. The association is linear and fairly strong. y x where y represents the number of days and x represents the temperature. For every 1 degree increase in average March temperature, in degrees
Celsius, we predict the number of days in April until first bloom to decrease by 4.69. The y -intercept is outside of the range of data and therefore has no meaningful interpretation. (c) Predicted number of days until 1 st bloom is ˆ 33.12 4.69(3.5) 16.7.
17th. (d) Predicted number of days until 1 st
We predict the first cherry blossom to appear on April
The observed value was 10. The residual is then 10 12.015
random fashion.
= − 2.015.
(e) The plot given below is the residuals vs. the explanatory variable. There is no discernable pattern in the residuals. They are clustered about 0 in a
(f) r 2 = 0.72
and s = 3.02.
72% of the variation in the number of days in April until the first cherry blossom appears is explained by the least-squares regression of the number of days in April until 1 st bloom on the average temperature, in Celsius, in March. We expect an average prediction error of 3.02 days.
80 The Practice of Statistics for AP* , 4/e

R3.6 (a) The slope of the regression line for predicting final-exam score from pre-exam totals is b = 0.6



8
30



= 0.16; for every extra point earned on the midterm, we predict that the score on the final exam will increase by about 0.16. The intercept of the regression line is a = − = (b)
(c) r 2 = 0.36, so only 36% of the variability in the final exam scores is accounted for by the linear relationship with pre-exam totals. About
64% of the individual variation is not accounted for by the least squares regression line, so Julie has a good reason to think this is not a good estimate.
R3.7 (a) If we left out Hawaii, the correlation would decrease because Hawaii is above average for both the maximum 24 hour precipitation and the maximum annual precipitation. (b) The blue line is the line calculated with all 50 states. Hawaii’s point is influential and pulls the line up toward it. The other line is the one with all states except Hawaii. (c) If we change the measurement on both x and y from inches to feet: the correlation will not change since it does not have units, s will decrease since it would now be measured in feet as well, the slope of the regression line would not change since both x and y are measured in the same units leading to a slope without units, but the y-intercept would decrease since it changes units from inches to feet. If we switch the explanatory and response variables, the correlation will not change, but the standard error and the least squares line will.
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T3.1 d. A correlation of near zero indicates no (or little) linear relationship, either positive or negative.
Answers a, b and e indicate some form of linear relationship. Answer c implies no relationship whatsoever. It is possible to have a correlation where there is a strong relationship, just not a linear one.
T3.2 e. This point is influential because it is well above the mean for the amount spent on tobacco and well below the mean for the amount spent on alcohol. The observation (4.5, 6.0) is not an outlier because it does not have the greatest value in either dimension, nor does it fall outside the main pattern of the data set.
T3.3 c. This is the definition of s x r
= = s s y x
which implies that
−
Looking at the residual plot, the fish with activity level 83 has a residual of about 3. Since residual = − ˆ , we find that r 2 .
T3.4 a. The slope for the least-squares line depends on which variable is the explanatory variable and which is the response. Also, the slope s y so b
ˆ residual 83 3 86.
y = = s y
> s x
.
T3.6 c. This is another way of saying the average error between the actual values and the predicted values using the linear model.
Chapter 3: Describing Relationships 81

T3.7 b. The correlation does not have units attached to it.
T3.8 e. Since s y s x
and both standard deviations are 1, the correlation will be the same as the slope.
T3.9 b. The slope gives the increase in y for each unit increase in x . So if x increases 5 units, then y units.
T3.10 c. Ethiopia is pulling the right hand side of the least-squares line up. If we remove that point, the line will be more shallow, resulting in a decrease in the slope and an increase in the y -intercept.
T3.11 (a) A scatterplot, with the regression line, is shown below.
(b) The regression line for predicting y = height from x y = + x (c) At age 480 cm. There are 2.54 cm to the inch, so her height in inches would be 255.934
= 100.76
in. (d) This height is impossibly large
(about 8 feet, 4 inches) because we used extrapolation. Obviously the linear trend does not continue all the way out to 40 years. Our data was based only on the first 5 years of life.
T3.12 (a) The unusual point is the one in the upper right hand corner with isotope value about 19.3
− and silicon value about 345. This point is unusual in that it has such a high silicon value for the given isotope value. (b) (i) If the point were removed the correlation would increase because this point does not follow the linear pattern of the other points. (ii) And since this point has a higher silicon value, if it were removed, the slope of the regression line would decrease and the y -intercept would increase. y The variable y represents the percent of the grass burned and the variable x represents the number of wildebeests. (b) The slope of the regression line suggests that for every increase of 1000 wildebeest (this is a 1 unit increase in x since x is measured in terms of 1000s of wildebeest), we predict the percent of grassy area burned will decrease by about 0.058.
(c) The overall pattern is moderately linear ( r = − 0.803).
(d) The linear model is appropriate for describing the relationship between wildebeest abundance and percent of grass area burned. The residual plot shows a fairly “random” scatter of points around the “residual = 0” line. There is one large positive residual at 1249 thousand wildebeest. Since r 2 = 0.646, 64.6% of the variation in percent of grass area burned is explained by the least-squares regression of percent of grass area burned on wildebeest abundance. That leaves 35.4% of the variation in percent of grass area burned unexplained by the linear relationship.
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